from station X a train starts from rest and attains a speed of 54km/h in 10s .Then by applying brakes negative acceleration of 2.5ms–²is produced and it stops at station Y in 6s.Find the distance between station X and Y.
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u is equal to zero speed is equal to 15mper second is equal to 10 second So v minus u/t is equal to a therefore 15 minus zero divided by 10 is equal to a therefore is equal to 1.5 m per second
now s equal to UT plus half a t square so S is equal to zero into 10 + half into 1.5 into 10 ka square s is equal to 75 again a is equal to 2.5 metre per second time is equal to 6 visual 20 so u is equal to 15 metre per second now is is equal to UT plus half a t square is equal to 135 metreshow distance from x to Y is equal to 135 minus 75 is equal to 60 m
now s equal to UT plus half a t square so S is equal to zero into 10 + half into 1.5 into 10 ka square s is equal to 75 again a is equal to 2.5 metre per second time is equal to 6 visual 20 so u is equal to 15 metre per second now is is equal to UT plus half a t square is equal to 135 metreshow distance from x to Y is equal to 135 minus 75 is equal to 60 m
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